Duo property for rings by the quasinilpotent perspective
aa r X i v : . [ m a t h . R A ] F e b DUO PROPERTY FOR RINGS BY THE QUASINILPOTENTPERSPECTIVE
ABDULLAH HARMANCI, YOSUM KURTULMAZ, AND BURCU UNGOR
Abstract.
In this paper, we focus on the duo ring property via quasinilpotentelements which gives a new kind of generalizations of commutativity. We callthis kind of ring qnil-duo . Firstly, some properties of quasinilpotents in a ringare provided. Then the set of quasinilpotents is applied to the duo propertyof rings, in this perspective, we introduce and study right (resp., left) qnil-duorings. We show that this concept is not left-right symmetric. Among othersit is proved that if the Hurwitz series ring H ( R ; α ) is right qnil-duo, then R isright qnil-duo. Every right qnil-duo ring is abelian. A right qnil-duo exchangering has stable range 1. Keywords:
Quasinilpotent element, duo ring, qnil-duo ring Introduction
Throughout this paper, all rings are associative with identity. Let N ( R ), J ( R ), U ( R ), C ( R ) and Id( R ) denote the set of all nilpotent elements, the Jacobson radical,the set of all invertible elements, the center and the set of all idempotents of a ring R , respectively. We denote the n × n full (resp., upper triangular) matrix ring over R by M n ( R ) (resp., U n ( R )), and D n ( R ) stands for the subring of U n ( R ) consistingof all matrices which have equal diagonal entries and V n ( R ) = { ( a ij ) ∈ D n ( R ) | a ij = a ( i +1)( j +1) for i = 1 , . . . , n − j = 2 , . . . , n − } is a subring of D n ( R ).Let Z and Z n denote the ring of integers and the ring of integers modulo n where n ≥ right (resp., left ) duo if every right (resp., left) ideal is an ideal, in other words, Ra ⊆ aR (resp., aR ⊆ Ra ) for every a ∈ R , and a ring is said to be duo if it is both rightand left duo. The duo ring property was studied in different aspects. For example,in [5], the concept of right unit-duo ring was introduced, namely, a ring R is called right unit-duo if for every a ∈ R , U ( R ) a ⊆ aU ( R ). Left unit-duo rings are definedsimilarly. In [9], the normal property of elements on Jacobson and nil radicals wereconcerned. A ring R is called right normal on Jacobson radical if J ( R ) a ⊆ aJ ( R ) for all a ∈ R . Left normal on Jacobson radical rings can be defined analogously.Also in [9], on the one hand, a ring R is said to satisfy the right normal on uppernilradical if N ∗ ( R ) a ⊆ aN ∗ ( R ) for all a ∈ R where N ∗ ( R ) is the upper nilradicalof R . Similarly, left normal on upper nilradical rings are defined similarly. Onthe other hand, a ring R is said to satisfy the right normal on lower nilradical if N ∗ ( R ) a ⊆ aN ∗ ( R ) for all a ∈ R where N ∗ ( R ) is the lower nilradical of R . Similarly,left normal on lower nilradical rings are defined similarly. Also, a ring R is called right nilpotent-duo if N ( R ) a ⊆ aN ( R ) for every a ∈ R . Left nilpotent duo ringsare defined similarly (see [7]).Motivated by the works on duo property for rings, the goal of this paper is toapproach the notion of duo rings by the way of quasinilpotent elements, in thisregard, we introduce the notion of qnil-duo rings. Firstly, we investigate someproperties of quasinilpotent elements which we need for the investigation of qnil-duo property. Then we study some properties of this class of rings and observethat being a qnil-duo ring need not be left-right symmetric. It is proved that anyright (resp., left) qnil-duo ring is abelian, and any exchange right (resp., left) qnil-duo ring has stable range 1. It is observed that regularity and strongly regularitycoincide for right (resp., left) qnil-duo rings. We also study on some extensions ofrings such as Dorroh extensions, Hurwitz series rings and some subrings of matrixrings in terms of qnil-duo property.2. Some properties of quasinilpotents
Let R be a ring and a ∈ R . The commutant and double commutant of a in R aredefined by comm( a ) = { b ∈ R | ab = ba } and comm ( a ) = { b ∈ R | bc = cb forall c ∈ comm( a ) } , respectively, and R qnil = { a ∈ R | ax is invertible in R forevery x ∈ comm( a ) } . Elements of the set R qnil are called quasinilpotent (see [6]).Note that J ( R ) = { a ∈ R | ax is invertible for x ∈ R } . If a ∈ N ( R ) and x ∈ comm( a ), then ax ∈ N ( R ) and 1 + ax ∈ U ( R ). So J ( R ) ⊆ R qnil , N ( R ) ⊆ R qnil and R qnil does not contain invertible elements, 0 ∈ R qnil but the identity is not in R qnil . In this section, we start to expose some properties of R nil and continue tostudy some other properties of quasinilpotent elements in rings. Example 2.1.
There are rings R such that J ( R ) is strictly contained in R qnil . Proof.
Let F be a field and R = M n ( F ) for some positive integer n . Then J ( R ) = 0and the matrix unit E n belongs to R qnil but not J ( R ). (cid:3) We now mention some of the known facts about quasinilpotents for an easy refer-ence.
UO PROPERTY FOR RINGS BY THE QUASINILPOTENT PERSPECTIVE 3
Proposition 2.2. (1)
Let R be a ring, a ∈ R and n a positive integer. If a n ∈ R qnil , then a ∈ R qnil , in particular every nilpotent element is in R qnil ([3, Proposition 2.7]) . (2) If R is a local ring, then U ( R ) ∩ R qnil = ∅ and R = U ( R ) ∪ R qnil ([3,Theorem 3.2]) . (3) Let R be a ring, a , b ∈ R . Then ab ∈ R qnil if and only if ba ∈ R qnil ([11,Lemma 2.2]) . (4) Let a ∈ R qnil and r ∈ U ( R ) . Then r − ar ∈ R qnil ([3, Lemma 2.3]) . (5) Let e = e ∈ ( R ) . Then ( eRe ) qnil = ( eRe ) ∩ R qnil ([16, Lemma 3.5]) . In the following, we determine quasinilpotent elements in some classes of rings.
Lemma 2.3.
Let R be a ring. Then the following hold. (1) (" a b c | a, c ∈ R qnil , b ∈ R ) ⊆ U ( R ) qnil . (2) (" a b a | a ∈ R qnil , b ∈ R ) ⊆ D ( R ) qnil . (3) Let A = " a b a ∈ D ( R ) qnil with b ∈ comm ( a ) . Then a ∈ R qnil .Proof. (1) Let a , c ∈ R qnil , b ∈ R , A = " a b c ∈ U ( R ) and B = " x y z ∈ comm( A ). Then AB = BA implies 1 − ax and 1 − cz are invertible. Hence I − AB is invertible. So A ∈ R qnil .(2) Let A = " a b a ∈ D ( R ) with a ∈ R qnil , b ∈ R and B = " x y x ∈ comm( A ).Then AB = BA implies x ∈ comm( a ). Then 1 − ax is invertible. Hence I − AB is invertible. So A ∈ D ( R ) qnil .(3) Clear. (cid:3) Proposition 2.4.
Let ( R i ) i ∈ I be a family of rings for some index set I and let R = Q i ∈ I R i . Then R qnil = Q i ∈ I R qnili .Proof. Let ( a i ), ( x i ) ∈ R . Then ( x i ) ∈ comm( a i ) if and only if x i ∈ comm( a i ) forall i ∈ I . Hence 1 + ( a i )( x i ) is invertible in R if and only if 1 + a i x i is invertible in R i for every i ∈ I . So the result follows. (cid:3) Let R be an algebra over a commutative ring S . The Dorroh extension (or idealextension ) of R by S denoted by I ( R, S ) is the direct product R × S with usualaddition and multiplication defined by ( a , b )( a , b ) = ( a a + b a + b a , b b )for a , a ∈ R and b , b ∈ S . ABDULLAH HARMANCI, YOSUM KURTULMAZ, AND BURCU UNGOR
Lemma 2.5.
Let I ( R, S ) be an ideal extension. Then the following hold. (1) For ( a, b ) ∈ I ( R, S ) , ( c, d ) ∈ comm ( a, b ) if and only if c ∈ comm ( a ) . (2) ( a, b ) has an inverse ( c, d ) in I ( R, S ) if and only if ( a + b )( c + d ) = 1 =( c + d )( a + b ) and bd = db = 1 .Proof. (1) ( c, d ) ∈ comm( a, b ) if and only if ( a, b )( c, d ) = ( c, d )( a, b ) if and only if ac + da + bc = ca + da + bc and bd = db if and only if ac = ca and bd = db if andonly if c ∈ comm( a ).(2) ( a, b )( c, d ) = (0 ,
1) = ( c, d )( a, b ) if and only if ac + da + bc = ca + da + bc = 0and bd = db = 1 if and only if ac + da + bc + bd + ( − bd ) = ( a + b )( c + d ) − bd = db = 1. (cid:3) Proposition 2.6.
Let I ( R, S ) be an ideal extension of R by S . Then (1) ( R, qnil = ( R, ∩ I ( R, S ) qnil . (2) (0 , S ) ∩ I ( R, S ) qnil ⊆ (0 , S ) qnil .Proof. (1) Let ( x, ∈ ( R, qnil and ( a, b ) ∈ I ( R, S ) with ( a, b ) ∈ comm( x, a ∈ comm( x ) and so 1 + xa is invertible in R . We prove (0 ,
1) + ( x, a, b ) isinvertible. Since S lies in the center of R , a + b ∈ comm( x ). Hence 1 + x ( a + b ) isinvertible, say (1 + x ( a + b ))( u + 1) = ( u + 1)(1 + x ( a + b )) = 1. This implies that u ( x ( a + b )) + x ( a + b ) + u = 0. Hence ((0 ,
1) + ( x, a, b ))( u,
1) = ( u, ,
1) +( x, a, b )) = (0 ,
1) for all ( a, b ) ∈ comm( x, x, ∈ I ( R, S ) qnil .Conversely, let ( x, ∈ ( R, ∩ I ( R, S ) qnil and ( r, ∈ comm( x, ,
1) + ( x, r,
0) = ( rx,
1) is invertible. Let ( a, b ) be the inverse of ( rx, rx, a, b ) = (0 ,
1) implies b = 1 and rxa + rx + a = 0. ( a, rx,
1) = (0 , arx + a + rx = 0. Hence (1 + a )(1 + rx ) = 1 and (1 + rx )(1 + a ) = 1.Hence (1 ,
0) + ( r, x,
0) is invertible in ( R,
0) for all ( r, ∈ comm( x, x, ∈ ( R, qnil or ( R, ∩ I ( R, S ) qnil ⊆ ( R, qnil .(2) Let (0 , s ) ∈ (0 , S ) ∩ I ( R, S ) qnil . Let (0 , b ) ∈ (0 , S ) with (0 , b ) ∈ comm(0 , s ). Then(0 ,
1) + (0 , s )(0 , b ) = (0 , sb ) is invertible in I ( R, S ). There exists ( u, v ) ∈ I ( R, S )such that (0 , sb )( u, v ) = ((1 + sb ) u, (1 + sb ) v ) = (0 ,
1) = ( u, v )(0 , sb ) =((1+ sb ) u, v (1+ sb )). Hence (1+ sb ) v = v (1+ sb ) = 1 and (1+ sb ) u = 0. Hence u = 0.Thus (0 , , s )(0 , b ) = (0 , sb ) is invertible in (0 , S ) with inverse (0 , v ) ∈ (0 , S ).It follows that (0 , s ) ∈ (0 , S ) qnil and so (0 , S ) ∩ I ( R, S ) qnil ⊆ (0 , S ) qnil . (cid:3) The following gives us necessary and sufficient conditions for (0 , S ) qnil to be con-tained in I ( R, S ) qnil . Theorem 2.7.
Let I ( R, S ) be the ideal extension of an algebra R by a commutativering S . Let (0 , i ) ∈ (0 , S ) qnil . Then (0 , i ) ∈ I ( R, S ) qnil if and only if for every UO PROPERTY FOR RINGS BY THE QUASINILPOTENT PERSPECTIVE 5 ( a, b ) ∈ comm (0 , i ) there exists ( u, v ) ∈ I ( R, S ) such that ( i ( a + b ) + 1)( u + v ) =(1 + ib ) v = 1 .Proof. Assume that (0 , i ) ∈ I ( R, S ) qnil . Let ( a, b ) ∈ comm(0 , i ) in I ( R, S ). Then(0 , , i )( a, b ) must be invertible. There exists ( u, v ) ∈ I ( R, S ) such that (0 ,
1) =((0 , , i )( a, b ))( u, v ). It follows that (0 ,
1) = ((0 , , i )( a, b ))( u, v ) = ( ia, ib )( u, v ) = ( iau + (1 + ib ) u + v ( ia ) , (1 + ib ) v ). Then iau + (1 + ib ) u + iav =0 and (1 + ib ) v = 1. They lead us (( i ( a + b ) + 1)( u + v ) = (1 + ib ) v . Hence( i ( a + b ) + 1)( u + v ) is invertible. Conversely, note that iau + (1 + ib ) u + iav = 0if and only if (( i ( a + b ) + 1)( u + v ) = (1 + ib ) v . Assume that for (0 , i ) ∈ (0 , S )there exists ( u, v ) ∈ I ( R, S ) such that ( i ( a + b ) + 1)( u + v ) = (1 + ib ) v = 1. Thenby concealing paranthesis we may reach that (0 ,
1) = ((0 ,
1) + (0 , i )( a, b ))( u, v ) for( a, b ) ∈ comm(0 , i ). Hence (0 , i ) ∈ I ( R, S ) qnil . (cid:3) Let R be a ring and S a subring of R with the same identity as that of R and T [ R, S ] = { ( r , r , · · · , r n , s, s, · · · ) : r i ∈ R, s ∈ S, n ≥ , ≤ i ≤ n } . Then T [ R, S ] is a ring under the componentwise addition and multiplication. Notethat N( T [ R, S ]) = T [ N ( R ) , N ( S )] and C ([ T, S ]) = T [ C ( R ) , C ( R ) ∩ C ( S )]. Proposition 2.8.
Let R be a ring and S a subring of R with the same identity asthat of R . (1) If A = ( a , a , a , . . . , a n , s, s, s, . . . ) ∈ T [ R, S ] qnil , then a i ∈ R qnil for i = 1 , , , . . . , n and s ∈ S qnil . (2) If a ∈ R qnil and s ∈ S qnil , then A = ( a, s, s, s, . . . , .. ) ∈ T [ R, S ] qnil .Proof. (1) Let A = ( a , a , a , . . . , a n , s, s, s, . . . ) ∈ T [ R, S ] qnil and b i ∈ comm( a i )and t ∈ comm( s ). Then B = ( b , b , b , . . . , b n , t, t, t, . . . ) ∈ comm( A ). Let 1 =(1 , , , . . . , , . . . ) denote the identity of T [ R, S ]. So 1 + AB is invertible. Therefore1 + a i b i is invertible for i = 1 , , , . . . , n and 1 + st is invertible in S . Hence a i ∈ R qnil for i = 1 , , , . . . , n and s ∈ S qnil .(2) Let a ∈ R qnil , s ∈ S qnil , A = ( a, s, s, . . . ). If B = ( b , b , . . . , b m , t, t, t, . . . . ) ∈ T [ R, S ] lies in comm( A ), then b ∈ comm( a ), b i ∈ comm( s ) for i = 2 , , . . . , m and t ∈ comm( s ). Hence 1+ ab and 1+ sb i are invertible in R where i = 2 , , . . . , m and1+ st is invertible in S . Hence 1+ AB is invertible in T [ R, S ]. So A = ( a, s, s, . . . ) ∈ T [ R, S ] qnil . (cid:3) Let R be a ring with an endomorphism α and let H ( R ; α ) be the set of formal expres-sions of the type f ( x ) = P ∞ n =0 a n x n where a n ∈ R for all n ≥
0. Define additionas componentwise and ∗ -product on H ( R ; α ) as follows: for f ( x ) = P ∞ n =0 a n x n ABDULLAH HARMANCI, YOSUM KURTULMAZ, AND BURCU UNGOR and g ( x ) = P ∞ n =0 b n x n , f ∗ g = P ∞ n =0 c n x n where c n = P nn =0 (cid:0) ni (cid:1) a i b n − i . Then H ( R ; α ) becomes a ring with identity containing R under these two operations.The ring H ( R ; α ) is called the Hurwitz series ring over R . The Hurwitz polynomialring h ( R ; α ) is the subring of H ( R ; α ) consisting of formal expressions of the form P ni =0 (cid:0) ni (cid:1) a i x i . Let ǫ : H ( R ; α ) → R defined by ǫ ( f ( x )) = a . Then ǫ is a homomor-phism with ker( ǫ ) = xH ( R ; α ) and H ( R ; α ) / ker( ǫ ) ∼ = R . There exist one-to-onecorrespondences between H ( R ; α ) and R relating to invertible elements, commu-tants and ideals. Let R [[ x ; α ]] be the skew formal power series ring over R . The sumis the same but multiplication in H ( R, α ) is similar to the usual multiplication of R [[ x ; α ]], except that binomial coefficients appear in each term in the multiplicationdefined in H ( R, α ). Also, there is a ring homomorphism ǫ between R [[ x ; α ]] and R ,defined by ǫ ( f ( x )) = a where f ( x ) = a + a x + a x + · · · ∈ R [[ x ; α ]]. Clearly, ǫ is an onto map and R [[ x ; α ]] / ker( ǫ ) ∼ = R . Lemma 2.9.
Let R be a ring and α a ring endomorphism of R . Then (1) U ( H ( R ; α )) = ǫ − U ( R ) , (2) U ( R [[ x ; α ]]) = ǫ − U ( R ) .Proof. It is routine. (cid:3)
In the next result, we determine the quasinilpotent elements of H ( R ; α ) and R [[ x ; α ]]. Proposition 2.10. (1)
Let H ( R ; α ) be a Skew Hurwitz series ring over R .Then H ( R ; α ) qnil = ǫ − R qnil . (2) Let R [[ x ; α ]] be a Skew formal power series ring over R . Then R ([[ x ; α ]]) qnil = ǫ − R qnil .Proof. (1) Let f ( x ) = a + a x + a x + · · · ∈ H ( R ; α ) qnil and r ∈ R with r ∈ comm R ( a ). Then r ∈ comm H ( R ; α ) ( a ). Then 1 + f ( x ) r ∈ U ( H ( R ; α )). Hence1 + a r ∈ U ( R ). So a ∈ R qnil . Since a = ǫ ( f ( x )), f ( x ) ∈ ǫ − ( R qnil ). Conversely,let g ( x ) = b + b x + b x + · · · ∈ ǫ − ( R qnil ). Then ǫ ( g ( x )) = b ∈ R qnil . Let h ( x ) = c + c x + c x + · · · ∈ comm H ( R,α ) ( g ( x )). Then c ∈ comm R ( b ) and so1 + b c ∈ U ( R ). So 1 + g ( x ) h ( x ) ∈ U ( H ( R, α )). Hence g ( x ) ∈ H ( R, α ) qnil . Thiscompletes the proof.(2) Similar that of (1). (cid:3) Qnil-duo Rings
In this section, we deal with the right duo property on the set of quasinilpotentelements. By this means, we give a generalization of commutativity from the per-spective of quasinilpotents.
UO PROPERTY FOR RINGS BY THE QUASINILPOTENT PERSPECTIVE 7
Definition 3.1.
A ring R is called right qnil-duo if R qnil a ⊆ aR qnil for every a ∈ R .Similarly, R is called left qnil-duo if aR qnil ⊆ R qnil a for every a ∈ R . If R is bothright and left qnil-duo, then it is called qnil-duo , i.e. R qnil a = aR qnil for every a ∈ R .The qnil-duo property of rings is not left-right symmetric as the following exampleshows. Example 3.2.
Let S = F ( t ) denote the quotient field of the polynomial ring F [ t ] over a field F and α : S → S defined by α ( f ( t ) /g ( t )) = f ( t ) /g ( t ). Let R = S [[ x ; α ]] denote the skew power series ring with xa = α ( a ) x for a ∈ S . Everyelement of R is of the form a = P ∞ i =0 a i x i . For any r = a + P ∞ i =1 a i x i with a = 0is invertible. Hence R qnil = xR . This ring is considered in [2, Lemma 1.3(3)],[8, Example 1] and in [7, Example 1.5]. As in the proof of [8, Example 1], for tx m ∈ tR qnil , there is no g ( x ) ∈ R qnil such that tx m = g ( x ) t . Hence R is not leftqnil-duo. We show that R is right qnil-duo. Let f ( x ) ∈ R qnil , g ( x ) ∈ R . We showthat there exists f ( x ) ∈ R qnil such that f ( x ) g ( x ) = g ( x ) f ( x ). Assume that g ( x )is invertible. Then f ( x ) g ( x ) = g ( x )( g ( x ) − f ( x ) g ( x ) ∈ g ( x ) R qnil , otherwise, let g ( x ) = h ( x ) x m where h ( x ) = a + a x + a x + · · · is invertible. Then f ( x ) g ( x ) = f ( x ) h ( x ) x m = f ( x ) x m h ( x ) = x m f ( x ) h ( x ) = x m h ( x )( h ( x ) − f ( x ) h ( x )) = g ( x )( h ( x ) − f ( x ) h ( x )) ∈ g ( x ) R qnil since f ( x ) is not invertible and f ( x ) is anapplication of x m to f ( x ) from the right, therefore f ( x ) = x k f ( x ) ∈ R qnil forsome k ≥
1, by Proposition 2.2(4), h ( x ) − f ( x ) h ( x ) ∈ R qnil . Thus R is rightqnil-duo. Examples 3.3. (1) All commutative rings, all division rings are qnil-duo.(2) There are local rings that are not right qnil-duo.
Proof. (1) When R is a commutative ring, it is both right and left qnil-duo. If R is a division ring, then R qnil = { } , therefore R is both right and left qnil-duo.(2) Let A = Z [ x, y ] be the polynomial ring with non-commuting indeterminates x and y and I be the ideal generated by the set { x , y , yx, x − xy, x − , x, y } .Consider the ring R = A/I . By [15, Example 7], R is a local ring. It is easilychecked that R qnil = { , , x, y, x, y, x + y, x + y } and ( R qnil ) = 0. 2 + x belongs to R qnil since it is nilpotent. For x ∈ R qnil and y ∈ R , xy ∈ R qnil y . It iseasily checked that there is no t ∈ R qnil such that xy = yt ∈ yR qnil . Hence R isnot right qnil-duo. (cid:3) Lemma 3.4.
Let R be a ring with R qnil central in R . Then R is qnil-duo. ABDULLAH HARMANCI, YOSUM KURTULMAZ, AND BURCU UNGOR
Proof.
Assume that R qnil is central in R . Let a ∈ R and b ∈ R qnil . Then b beingcentral implies ab = ba ∈ aR qnil . (cid:3) Theorem 3.5.
Let { R i } i ∈ I be a family of rings for some index set I and R = Q i ∈ I R i . Then R i is right (resp., left) qnil-duo for each i ∈ I if and only if R isright (resp., left) qnil-duo.Proof. Assume that R i is right (resp., left) qnil-duo for each i ∈ I . Let a = ( a i ) ∈ R , b = ( b i ) ∈ R qnil . By Proposition 2.4, b i ∈ R qnili for each i ∈ I . By assumptionthere exists c i ∈ R qnili such that b i a i = a i c i for each i ∈ I . Set c = ( c i ). Then ba = ac ∈ aR qnil . Hence R qnil a ⊆ aR qnil . Conversely, suppose that R is rightqnil-duo. Let a i ∈ R i and b i ∈ R qnili where i ∈ I . Let a = ( a i ), b = ( b i ) ∈ R where i th -entry of a is a i and the other entries are 0 and i th -entry of b is b i and the otherentries are 0, respectively. Then a = ( a i ) ∈ R and by Proposition 2.4, b ∈ R qnil .Supposition implies there exists c = ( c i ) ∈ R qnil such that ba = ac . Comparingsentries of both sides we have b i a i = a i c i . By Proposition 2.4, c i ∈ R qnili . Thus foreach i ∈ I , R i is right qnil-duo. Similarly, it is proven that for each i ∈ I , R i is leftqnil-duo. (cid:3) Recall that a ring R is called abelian if every idempotent in R is central. Theorem 3.6.
Let R be a ring. Then the following hold. (1) ex − exe and xe − exe ∈ R qnil for every x and e = e ∈ R . (2) Right (resp., left) qnil-duo rings are abelian. (3)
Let R be a ring and e ∈ Id ( R ) . If R is right (resp., left) qnil-duo ring, then eR and (1 − e ) R are right (resp., left) qnil-duo rings. The converse holdsif e is central.Proof. (1) Let t ∈ comm( ex − exe ). Then t ( ex − exe ) = ( ex − exe ) t . So we have( t ( xe − exe )) = 0. Hence 1 − ( ex − exe ) t is invertible and so ex − exe ∈ R qnil .Similarly, xe − exe ∈ R qnil .(2) Let e = e ∈ R . By hypothesis, R qnil e ⊆ eR qnil . By (1), xe − exe ∈ R qnil for all x ∈ R . It implies for any x ∈ R , there exists t ∈ R qnil such that ( xe − exe ) e = et .Multiplying the latter equality by e from the left we have et = 0. So xe = exe .Similarly, ex = exe since ex − exe ∈ R qnil by (1). Hence R is abelian.(3) It is clear by Theorem 3.5. (cid:3) Corollary 3.7.
Let R be a right (resp., left) qnil-duo ring and e ∈ Id ( R ) . Thenthe corner ring eRe is a right (resp., left) qnil-duo ring.Proof. The ring R being right (resp., left) qnil-duo, e is central in R by Theorem3.6(2). Hence Theorem 3.6(3) completes the proof. (cid:3) UO PROPERTY FOR RINGS BY THE QUASINILPOTENT PERSPECTIVE 9
Theorem 3.8.
Every right (resp., left) qnil-duo ring is directly finite.Proof.
Let R be a right qnil-duo ring and a , b ∈ R with ab = 1. Set e = 1 − ba .Then e is an idempotent. By Theorem 3.6, e is central. So 0 = ae = ea . Hence0 = a − ba . Multiplying the latter by b from the right, we get 1 = ba . (cid:3) There is a directly finite ring that is neither right nor left qnil-duo.
Example 3.9.
Consider the ring R = M ( Z ). Then R is a directly finite ring butnot abelian. Hence it is neither right nor left qnil-duo.We apply Theorem 3.6 to show that full matrix rings and upper triangular matrixrings need not be right (resp., left) qnil-duo. But there are some subrings of fullmatrix rings that are qnil-duo. Examples 3.10. (1) For any ring R and any positive integer n , M n ( R ) and U n ( R ) are neither right nor left qnil-duo.(2) If R is commutative, then V n ( R ) is qnil-duo.(3) V n ( R [[ x ; σ ]]) is neither right nor left qnil-duo. Proof. (1) The rings M n ( R ) and U n ( R ) are not abelian. By Theorem 3.6(2), theyare neither right nor left qnil-duo.(2) If R is a commutative ring, V n ( R ) is also commutative, therefore it is right andleft qnil-duo.(3) Let R be a ring with an endomorphism σ . Assume that there exists a ∈ R suchthat σ ( a ) / ∈ a R . Let A = x x x x x x ∈ V ( R [[ x ; σ ]]) qnil , B = a a a a a a ∈ V ( R [[ x ; σ ]]). Assume that there exists D ∈ V ( R [[ x ; σ ]]) qnil such that AB = BD .Then (1 ,
1) entry of AB is σ ( a ) x and that of BD is a xf ( x ) for some f ( x ) ∈ R [[ x ; σ ]]. This contradicts with the choice of σ and a . Therefore V ( R [[ x ; σ ]]) isnot right qnil-duo. Similarly, it can be shown that V ( R [[ x ; σ ]]) is not left qnil-duo. (cid:3) Theorem 3.11.
Let R be a local ring with ( R qnil ) = 0 . Then R is right (resp.,left) qnil-duo.Proof. By Proposition 2.2(2), we have R = U ( R ) ∪ R qnil . We prove R qnil a ⊆ aR qnil .Let a ∈ R , b ∈ R qnil . If ba = 0, then we are done since ba = 0 = a ∈ aR qnil .Otherwise, i.e., if ba = 0, then we divide the proof in some cases.Case I. Let a / ∈ R qnil . Then a ∈ U ( R ). By Proposition 2.2(4), a − ba ∈ R qnil since b ∈ R qnil . Then ba = a ( a − ba ) ∈ aR qnil . Case II. Let a ∈ R qnil . By hypothesis, ba = 0, this contradicts with ba = 0.Therefore R is right qnil-duo ring. Similarly, we may prove aR qnil ⊆ R qnil a foreach a ∈ R . (cid:3) As an illustration of Theorem 3.11, we give the following examples. Also, thecondition ( R qnil ) = 0 in Theorem 3.11 is not superfluous. Example 3.12. (1) Consider the ring R = a b c a
00 0 a ∈ D ( Z ) . Then R qnil = a b c a
00 0 a ∈ R | a ∈ Z , b, c ∈ Z . So ( R qnil ) = 0. By Theorem3.11, R is qnil-duo.(2) Let R denote the ring in Examples 3.3. Then R qnil = { , , x, y, x, y, x + y, x + y } and ( R qnil ) = 0 and 2 + x belongs to R qnil since it is nilpotentand (2 + x ) = 0. Since R is local and R qnil does not contain invertible elements, R qnil = J ( R ). To complete the proof we may assume that x , y ∈ R qnil . Then xy = 2 and xy ∈ R qnil y . It is easily checked that there is no t ∈ R qnil such that xy = yt ∈ yR qnil . Hence R is not right qnil-duo. Compare to Theorem 3.11.Note that by Theorem 3.11, if R is a division ring, D ( R ) is a qnil-duo ring. Onemay ask whether D ( R ) is qnil-duo over a domain R . The following exampleanswers negatively. Example 3.13.
Consider the ring D ( R [[ x ]]) in [9, Example 1.4(1)]. It is provedthat D ( R [[ x ]]) is neither right nor left normal property of elements on Jacobsonradical. Since J ( D ( R [[ x ]])) = D ( R [[ x ]]) qnil , D ( R [[ x ]]) is neither right nor leftqnil-duo. Theorem 3.14.
Let R be a domain. If D ( R ) is right (resp., left) qnil-duo, then R is right (resp., left) qnil-duo.Proof. Assume that D ( R ) is right qnil-duo. Let a ∈ R and b ∈ R qnil . Consider A = " a a = 0, B = " b b = 0. Let X = " x y x ∈ comm( B ). Then I − BX is invertible since 1 − bx is invertible in R . Hence BA ∈ D ( R ) qnil A . There exists C = " c d c ∈ D ( R ) qnil such that BA = AC . Then ba = ac and ad = 0. Byhypothesis d = 0. By Lemma 2.3(3), c ∈ R qnil . It follows that ba = ac ∈ aR qnil .Hence R qnil a ⊆ aR qnil . (cid:3) UO PROPERTY FOR RINGS BY THE QUASINILPOTENT PERSPECTIVE 11
Recall that a ring R is said to have stable range 1 if for any a , b ∈ R satisfying aR + bR = R , there exists y ∈ R such that a + by is right invertible (cf. [14]). In[12], a ring R is called exchange if for any x ∈ R , there exists e ∈ Id( R ) such that e ∈ Rx and 1 − e ∈ R (1 − x ), and it is proved that for an abelian ring R , R isexchange if and only if it is clean, and R is exchange if and only if idempotents liftmodulo every left (or right) ideal. Theorem 3.15.
The following hold. (1)
Right (resp., left) qnil-duo exchange rings have stable range 1. (2)
Right (resp., left) qnil-duo regular rings (in the sense of von Neumann) arestrongly regular.Proof. (1) Let R be a right qnil-duo exchange ring. By Theorem 3.6, R is abelian.Hence [17, Theorem 6] implies R has stable range 1.(2) Let R be a qnil-duo regular ring and a ∈ R . There exists b ∈ R such that a = aba . Then ab = ( ab ) , ba = ( ba ) ∈ Id( R ). By Theorem 3.6, ab is central. So a = aba = a b . Hence R is strongly regular. (cid:3) Let R be a ring. The Jacobson radical of the polynomial ring R [ x ] is J ( R [ x ]) = N [ x ]where N = J ( R [ x ]) ∩ R is a nil ideal of R . Then N ⊆ R qnil and J ( R [[ x ]]) = xR [[ x ]].Therefore, R [[ x ]] qnil = xR [[ x ]]. One may wonder whether or not R [ x ] and R [[ x ]]are qnil-duo. The following example shows that R [ x ] and R [[ x ]] need not be rightqnil-duo. Example 3.16. (1) Let F be a field, R = M n ( F ) and consider the ring R [ x ].Observe that M n ( F [ x ]) is not right (or left) qnil-duo for any positive integer n ≥ R [ x ] is not right (or left) qnil-duo since M n ( F )[ x ] ∼ = M n ( F [ x ]).(2) Let R = A/ ( ab − ba −
1) denote the Weyl algebra discussed in [9, Example1.2(2)]. Let S = R [[ x ]]. Then S qnil = xR [[ x ]] = J ( R [[ x ]]), R is a domain and R [[ x ]]is abelian. It is proved that R [[ x ]] is neither right normal nor left normal on J ( R ).Therefore, R [[ x ]] is neither right qnil-duo nor left qnil-duo. Theorem 3.17.
Let R be an algebra over a commutative ring S . Consider theDorroh extension (or ideal extension) I ( R, S ) of R by S . If I ( R, S ) is right qnil-duo, then so is R .Proof. Assume that I ( R, S ) is right qnil-duo. Let a ∈ R , b ∈ R qnil . Then ( a, ∈ I ( R, S ) and ( b, ∈ I ( R, S ) qnil . Indeed, let ( x, y ) ∈ comm( b, x ∈ comm( b ). Since R is an algebra over S , we have x + y ∈ comm( b ). Then 1+ b ( x + y ) isinvertible in R with inverse t . Again by Lemma 2.5, (0 , b, x, y ) is invertible in I ( R, S ) with the inverse ( t − , b, a, ∈ I ( R, S ) qnil ( a, c, s ) ∈ I ( R, S ) qnil such that ( b, a,
0) = ( a, c, s ). So ba = a ( c + s ). To completethe proof we show c + s ∈ R qnil . Let x ∈ comm( c + s ). Then cx + sx = sc + xs .Since R is an algebra over S , sx = xs , this implies cx = xc , and so x ∈ comm( c ).Hence ( x, ∈ comm( c, s ). Since ( c, s ) ∈ I ( R, S ) qnil , (0 ,
1) + ( c, s )( x,
0) is invertiblein I ( R, S ). Thus 1 + ( c + s ) x is invertible in R by Lemma 2.5(2). So c + s ∈ R qnil .Therefore R is right qnil-duo. (cid:3) Proposition 3.18.
Let R be a ring and S a subring of R . If T [ R, S ] is rightqnil-duo, then so are R and S . The converse holds if S qnil ⊆ R qnil .Proof. Assume that T [ R, S ] is right qnil-duo. Let a ∈ R , b ∈ R qnil . Let A =( a, , , , . . . ), B = ( b, , , , . . . ) ∈ T [ R, S ]. By Proposition 2.8, B ∈ T [ R, S ] qnil .By supposition there exists C = ( c , c , · · · , c m , t, t, . . . . ) ∈ T [ R, S ] qnil such that BA = AC . Hence ba = ac . By Proposition 2.8, c ∈ R qnil . Similarly, let s ∈ S , t ∈ S qnil and C = (0 , s, s, s, s, . . . , s, . . . ), D = (0 , t, t, t, . . . . ) ∈ T [ R, S ]. By Proposition2.8, D ∈ T [ R, S ] qnil . There exists D ′ = ( d , d , d , . . . ., d l , u, u, u, . . . ) ∈ T [ R, S ] qnil such that DC = CD ′ . By Proposition 2.8, u ∈ S qnil and ts = su ∈ sS qnil .Suppose that R and S are right qnil-duo and S qnil ⊆ R qnil . Let A ∈ T [ R, S ], B ∈ T [ R, S ] qnil where A = ( a , a , · · · , a n , s, s, · · · ), B = ( b , b , . . . , b m , t, t, . . . ),we prove BA = AC for some C ∈ T [ R, S ] qnil . By Proposition 2.8, b i ∈ R qnil for i = 1 , , . . . , m and t ∈ S qnil . By supposition b i ∈ R qnil implies b i a i = a i c i for some c i ∈ R qnil . We divide the proof in some cases:Case I. n ≤ m . Then b i a i ∈ R qnil a i . Since R is right qnil-duo, there exist c i ∈ R qnil such that b i a i = a i c i for each 1 ≤ i ≤ n . For n + 1 ≤ i ≤ m , b i s ∈ R qnil s . Thereexist c i ∈ R qnil such that b i s = sc i . For ts ∈ S qnil s , there exists l ∈ S qnil suchthat ts = sl ∈ sS qnil . Let C = ( c , c , c , . . . , c m , l, l, l, . . . ). By Proposition 2.8(2), C ∈ T [ R, S ] qnil . Then BA = AC ∈ AT [ R, S ] qnil .Case II. n > m . Let 1 ≤ i ≤ m . Then b i a i ∈ R qnil a i and since R is right qnil-duo, there exist c i ∈ R qnil such that b i a i = a i c i . For m + 1 ≤ i ≤ n , ta i ∈ S qnil a i . By S qnil ⊆ R qnil , we have ta i = a i c i ∈ a i R qnil for some c i ∈ R qnil . For ts ∈ S qnil s , by supposition there exists l ∈ S qnil such that ts = sl ∈ sS qnil . Let C = ( c , c , c , . . . , c n , l, l, l, . . . ). By Proposition 2.8(2), C ∈ T [ R, S ] qnil . Then BA = AC . Hence T [ R, S ] qnil A ⊆ AT [ R, S ] qnil . It completes the proof. (cid:3) Theorem 3.19. (1)
Let H ( R ; α ) be a Skew Hurwitz series ring over a ring R .If H ( R ; α ) is right qnil-duo, then R is right qnil-duo. (2) Let R [[ x ; α ]] be a skew formal power series ring over a ring R . If R [[ x ; α ]] is right qnil-duo, then R is right qnil-duo. UO PROPERTY FOR RINGS BY THE QUASINILPOTENT PERSPECTIVE 13
Proof.
Suppose that H ( R ; α ) is a right qnil-duo ring. Let a ∈ R qnil and b ∈ R .By the definition of ǫ and Proposition 2.10, there exist f ( x ), g ( x ) ∈ H ( R ; α ) with f ( x ) ∈ H ( R ; α ) qnil and ǫ ( f ( x )) = a , ǫ ( g ( x )) = b . There exists h ( x ) = c + c x + c x + · · · ∈ H ( R ; α ) qnil such that f ( x ) g ( x ) = g ( x ) h ( x ). Hence ǫ ( f ( x ) g ( x )) = ǫ ( g ( x ) h ( x )) implies ab = bc ∈ bR qnil . Thus R qnil b ⊆ bR qnil .(2) Similar to that of (1). (cid:3) Some subrings of matrix rings
Besides, for any ring R and any positive integer n ≥ M n ( R ) is not right (or left)qnil-duo, in this section, quasinilpotent elements of some subrings of full matrixrings are determined for the purpose of the use whether or not their subrings to beright (or left) qnil-duo. The rings L ( s,t ) ( R ) : Let R be a ring and s , t ∈ C ( R ).Let L ( s,t ) ( R ) = a sc d te f ∈ M ( R ) | a, c, d, e, f ∈ R , where the operationsare defined as those in M ( R ). Then L ( s,t ) ( R ) is a subring of M ( R ). Lemma 4.1.
Let A = a sc d te f ∈ L ( s,t ) ( R ) . Then the following hold. (1) A is invertible in L ( s,t ) ( R ) if and only if a , d and f are invertible in R . (2) If a , d , f ∈ R qnil , then A ∈ L ( s,t ) ( R ) qnil .Proof. (1) One way is clear. Let A = a sc d te f ∈ L ( s,t ) ( R ). Assume that a , d and f are invertible with ax = xa = 1, dz = zd = 1 and f v = vf = 1 where x, z, v ∈ R . Consider B = x sy z tu v ∈ L ( s,t ) ( R ) where y = − zcx and u = − zev .Then AB = BA = I .(2) Assume that a , d , f ∈ R qnil . We prove that A ∈ L ( s,t ) ( R ) qnil . Let B = x sy z tu v ∈ L ( s,t ) ( R ) with B ∈ comm( A ). It is easily checked that x ∈ comm( a ), z ∈ comm( d ), v ∈ comm( f ). Then 1 + ax , 1 + dz , 1 + f v are in-vertible in R . By (1), I + AB = ax scx + sdy dz tdu + tev f v is invertible. So A ∈ L ( s,t ) ( R ) qnil . (cid:3) Lemma 4.2.
Let A = a sc d te f ∈ L ( s,t ) ( R ) . Then the following hold. (1) If A ∈ L (0 ,t ) ( R ) qnil , then a ∈ R qnil . (2) If A ∈ L ( s, ( R ) qnil , then f ∈ R qnil . (3) A ∈ L (0 , ( R ) qnil if and only if a, d, f ∈ R qnil .Proof. (1) Let A = a d te f ∈ L (0 ,t ) ( R ) qnil and x ∈ comm( a ). Consider B = x ∈ L (0 ,t ) ( R ). Then B ∈ comm( A ). Since A ∈ L (0 ,t ) ( R ) qnil , I + AB isinvertible in L (0 ,t ) ( R ). By Lemma 4.1(1), 1 + ax ∈ U ( R ). Therefore a ∈ R qnil .(2) Similar to the proof of (1).(3) The sufficiency follows from Lemma 4.1(2). For the necessity, a, f ∈ R qnil by (1)and (2), respectively. Also, by the similar discussion in (1), we obtain d ∈ R qnil . (cid:3) Theorem 4.3.
Let R be a ring. If L (0 ,t ) ( R ) is right qnil-duo, then R is a rightqnil-duo ring.Proof. Assume that L (0 ,t ) ( R ) is right qnil-duo and let a ∈ R and b ∈ R qnil .Consider A = a , B = b ∈ L (0 ,t ) ( R ). By Lemma 4.1, B ∈ L (0 ,t ) ( R ) qnil . By supposition there exists B ′ = x z tu v ∈ L (0 ,t ) ( R ) qnil suchthat BA = AB ′ . It implies ba = ax . By Lemma 4.2(1), x ∈ R qnil . Hence ba = ax ∈ aR qnil . Thus R qnil a ⊆ aR qnil . (cid:3) There are right qnil-duo rings R such that the rings L ( s,t ) ( R ) need not be rightqnil-duo as shown below. Example 4.4.
The ring L (1 , ( Z ) is not right qnil-duo. UO PROPERTY FOR RINGS BY THE QUASINILPOTENT PERSPECTIVE 15
Proof.
Let A = ∈ L (1 , ( Z ) and B = ∈ L (1 , ( Z ) qnil .Assume that there exists C = x y z v u ∈ L (1 , ( Z ) qnil such that BA = AC .Then BA = and AC = x + 2 y z v + u u . BA = AC implies3 = 2 v + u and 2 = 3 u . These equations lead us a contradiction. Hence L (1 , ( Z )is not right qnil-duo. (cid:3) The rings H ( s,t ) ( R ) : Let R be a ring and s , t ∈ C ( R ) be invertible in R . Let H ( s,t ) ( R ) = a c d e f ∈ M ( R ) | a, c, d, e, f ∈ R, a − d = sc, d − f = te .Then H ( s,t ) ( R ) is a subring of M ( R ). Lemma 4.5.
Let A = a c d e f , B = x y z u v ∈ H ( s,t ) ( R ) . Then (1) AB = BA if and only if ax = xa , dz = zd , f v = vf . (2) A is invertible with inverse B if and only if ax = xa = 1 , dz = zd = 1 , f v = vf = 1 . (3) A ∈ H ( s,t ) ( R ) qnil if and only if a , d , f ∈ R qnil .Proof. (1) The necessity is clear. For the sufficiency, suppose that ax = xa , dz = zd , f v = vf . The matrix AB has cx + dy as (2 ,
1) entry, du + ev as (2 ,
3) entry and BA has ya + zc as (2 ,
1) entry, ze + uf as (2 ,
3) entry. To show AB = BA it is enoughto get cx + dy = ya + zc and du + ev = ze + uf . Now scx + sdy = ax + d ( sy − x ) = ax − dz = xa − za + za − dz = sya + szc . So cx + dy = ya + zc since s is invertible.Similarly, we get du + ev = ze + uf .(2) One way is clear. Assume that ax = xa = 1, dz = zd = 1, f v = vf = 1. Let B ∈ H ( s,t ) ( R ) with y = − zcx and u = − zev . Then AB = BA = I .(3) Assume that A ∈ H ( s,t ) ( R ) qnil . Let x ∈ comm( a ), y ∈ comm( f ). Let D = x s − x − t − y y . Then D ∈ comm( A ). In fact, scx = ( a − d ) x and tey = ( d − f ) y . Hence I + AD is invertible in H ( s,t ) ( R ). It follows that 1 + ax , 1 + f y ∈ U ( R ). So a, f ∈ R qnil . As for d ∈ R qnil , let r ∈ comm( d ) and D = − s − r r t − r .Then D ∈ comm( A ). By assumption I + AD ∈ U ( H ( s,t ) ( R )). Hence 1+ dr ∈ U ( R ).Hence d ∈ R qnil . Conversely, suppose that a , d , f ∈ R qnil . Let B ∈ comm( A ).Then x ∈ comm( a ), z ∈ comm( d ) and v ∈ comm( f ). By supposition, 1 + ax ,1 + dy and 1 + f v are invertible. By part (2), I + AB ∈ U ( H ( s,t ) ( R )). Hence A ∈ H ( s,t ) ( R ) qnil . This completes the proof. (cid:3) Theorem 4.6.
Let R be a ring. Then R is right qnil-duo if and only if H ( s,t ) ( R ) is right qnil-duo.Proof. Assume that R is a right qnil-duo ring. Let A = a c d e f ∈ H ( s,t ) ( R )and B = x y z u v ∈ H ( s,t ) ( R ) qnil . By Lemma 4.5, x , z , v ∈ R qnil . There exist x ′ , z ′ , v ′ ∈ R qnil such that xa = ax ′ , zd = dz ′ , vf = f v ′ . Let y ′ = s − ( x ′ − z ′ ) and u ′ = t − ( z ′ − v ′ ) and B ′ = x ′ y ′ z ′ u ′ v ′ . Then B ′ ∈ H ( s,t ) ( R ) qnil . We next showthat BA = AB ′ . It is enough to see ya + zc = cx ′ + dy ′ and ze + uf = du ′ + ev ′ . Westart with, cx ′ + dy ′ = cx ′ + ds − x ′ − ds − z ′ . Multiplying the latter from the leftby s and using the fact that s is central, we have s ( cx ′ + dy ′ ) = scx ′ + dx ′ − dz ′ =( sc + d ) x ′ − zd = ax ′ − zd = xa − zd = ( xa − za )+( za − zd ) = sya + szc = s ( ya + zc ).Since s is invertible, ya + zc = cx ′ + dy ′ . Similarly, du ′ + ev ′ = dt − z ′ − dt − v ′ + ev ′ .Multiplying the latter from the left by t and using the fact that t is central, wehave t ( du ′ + ev ′ ) = dz ′ − dv ′ + tev ′ = zd + ( te − d ) v ′ = zd − f v ′ = zd − vf = zd − zf + zf − vf = z ( d − f ) + ( z − v ) f = t ( ze + uf ). By using invertibility of t , weget du ′ + ev ′ = ze + uf . Conversely, suppose that H ( s,t ) ( R ) is a right qnil-duo ring.Let a ∈ R and b ∈ R qnil . Consider A = aI , B = bI ∈ H ( s,t ) ( R ). By Lemma 4.5, B ∈ H ( s,t ) ( R ) qnil . By supposition there exists B ′ = x y z u v ∈ H ( s,t ) ( R ) qnil such that BA = AB ′ . It implies ba = ax . Again by Lemma 4.5, x ∈ R qnil . Hence ba = ax ∈ aR qnil . Thus R qnil a ⊆ aR qnil . (cid:3) UO PROPERTY FOR RINGS BY THE QUASINILPOTENT PERSPECTIVE 17
Generalized matrix rings:
Let R be a ring and s ∈ U ( R ). Then " R RR R becomes a ring denoted by K s ( R ) with addition defined componentwise and mul-tiplication defined in [10] by " a x y b a x y b = " a a + sx y a x + x b y a + b y sy x + b b . In [10], K s ( R ) is called a generalized matrix ring over R . Lemma 4.7.
Let R be a ring. Then the following hold. (1) U ( K ( R )) = (" a bc d ∈ K ( R ) | a, d ∈ U ( R ) ) . (2) C ( K ( R )) = (" a a ∈ K ( R ) | a ∈ C ( R ) ) .Proof. (1) Let A = " a bc d ∈ U ( K ( R )). There exists B = " x yz t ∈ K ( R ) suchthat AB = BA = I , where I is the identity matrix. Then we have ax = xa = 1and dt = td = 1 . So a and d are invertible. Conversely, let A = " a bc d ∈ K ( R )with a , d ∈ U ( R ). Let x = a − , t = d − , k = − a − bd − and l = − d − ca − . Then B = " x kl t is the inverse of A in K ( F ).(2) Let A = " a bc d ∈ C ( K ( R )). By commuting A in turn with the matrices " and " in K ( R ) we reach at A = " a a . For the converse, let A = " a a ∈ K ( R ) where a ∈ C ( R ). Then clearly, A commutes with everyelement of K ( R ). So A ∈ C ( K ( R )). (cid:3) Proposition 4.8.
Let R be a ring and A = " a bc d ∈ K ( R ) . If a , d ∈ R qnil , then A ∈ K ( R ) qnil .Proof. Suppose that a , d ∈ R qnil . Let B = " x yz t ∈ K ( R ) with B ∈ comm( A ).Then x ∈ comm( a ), t ∈ comm( d ). Let r = 1 + ax , v = 1 + dt , s = ay + bt and u = cx + dz . By assumption, r = 1 + ax and v = 1 + dt are invertible in R . Let k = − r − sv − and l = − v − ur − . Then I + AB = " r su v is invertible with theinverse C = " r − kl v − . (cid:3) Let R be a ring, a , b ∈ R . Define l a − r b : R → R by ( l a − r b )( r ) = ar − rb and l b − r a : R → R by ( l b − r a )( r ) = br − ra . In [1], a local ring R is called bleached iffor any a ∈ J ( R ) and any b ∈ U ( R ), the abelian group endomorphisms l b − r a and l a − r b of R are surjective. Such rings is called uniquely bleached if the appropriatemaps are injective as well as surjective. In [13], R is a weakly bleached ring providedthat for any a ∈ J ( R ), b ∈ J ( R ), l a − l b and l b − l a are surjective and it is provedthat matrices over 2-projective free rings are strongly J-clean. It is proved thatall upper triangular matrices over bleached local rings are strongly clean. In [12,Example 2] and [1, Theorem 18] it is proved that a local ring R is weakly bleachedif and only if the 2 × U ( R ) is strongly clean. In thepreceding, maps of the form l a − r b play a central role. In this vein, we make use ofthe abelian group endomorphisms l a − r b to get the following result as partly theconverse of Proposition 4.8. Theorem 4.9.
Let R be a ring and A = " a bc d ∈ K ( R ) qnil . If for any x ∈ comm ( a ) and y ∈ comm ( d ) and for the abelian group endomorphisms l y − r x and l x − r y , b ∈ Ker ( l x − r y ) and c ∈ Ker ( l y − r x ) , then a , d ∈ R qnil .Proof. Assume that A = " a bc d ∈ K ( R ) qnil , x ∈ comm( a ) and y ∈ comm( d ), for l y − r x and l x − r y , b ∈ Ker( l x − r y ) and c ∈ Ker( l y − r x ). Then b ∈ Ker( l x − r y )implies ( l x − r y )( b ) = 0. So xb = by . c ∈ Ker( l y − r x ) implies ( l y − r x )( c ) = 0. So yc = cx . Let B = " x y ∈ K ( R ). Then xb = by and yc = xc give rise to B ∈ comm( A ). By hypothesis, I + AB is invertible. Then Lemma 4.7 implies 1 − ax and 1 − dy are invertible. Hence a , d ∈ R qnil . (cid:3) We may determine the set K ( R ) qnil for some rings R . Proposition 4.10. (1) If R is a local ring, then A = " a bc d ∈ K ( R ) qnil ifand only if a , d ∈ R qnil . (2) Let R be a ring. Then A = " a d ∈ K ( R ) qnil if and only if a , d ∈ R qnil . UO PROPERTY FOR RINGS BY THE QUASINILPOTENT PERSPECTIVE 19
Proof. (1) Assume that R is a local ring and A = " a bc d ∈ K ( R ) qnil , and d / ∈ R qnil . By Proposition 2.2, d ∈ U ( R ). In this case, 1+ d can not belong to U ( R ). ByLemma 4.7, I + A can not belong to U ( K ( R )). This contradicts A ∈ K ( R ) qnil .It follows that d ∈ R qnil . Similarly, we obtain a ∈ R qnil . The converse is clear byProposition 4.8.(2) Clear. (cid:3) There are some classes of rings R in which K ( R ) being a right qnil-duo ring implies R being a right qnil-duo ring. Theorem 4.11.
Let R be a ring. Then K ( R ) being a right qnil-duo ring implies R being a right qnil-duo ring if R is one of the following rings. (1) R is local. (2) R has no nonzero zero divisors.Proof. (1) Let R be a local ring. Assume that K ( R ) is a right qnil-duo ring. Let a ∈ R , b ∈ R qnil . Consider A = " a a , X = " b b ∈ K ( R ). By Proposition4.8, X ∈ K ( R ) qnil . There exists X ′ = " x ′ y ′ z ′ t ′ ∈ K ( R ) qnil such that XA = AX ′ .Hence ba = ax ′ . By Proposition 4.10, x ′ ∈ R qnil . So ba = ax ′ ∈ aR qnil .(2) Let R be a ring having no nonzero zero divisors. Assume that K ( R ) is a rightqnil-duo ring. Let a ∈ R , b ∈ R qnil . If a = 0 or b = 0, there is nothing to do. Let a = 0 and b = 0 and consider A = " a a , B = " b b ∈ K ( R ). By Proposition4.8, B ∈ K ( R ) qnil . There exists B ′ = " x ′ y ′ z ′ t ′ ∈ K ( R ) qnil such that BA = AB ′ .It implies ba = ax ′ = at ′ , ay ′ = 0 and az ′ = 0. Hence x ′ = t ′ and y ′ = z ′ = 0.Hence x ′ ∈ R qnil by Proposition 4.10. (cid:3) References [1] G. Borooah, A. J. Diesl and T. J. Dorsey
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Abdullah Harmanci, Department of Mathematics, Hacettepe University, Ankara, Turkey
Email address : [email protected] Yosum Kurtulmaz, Department of Mathematics, Bilkent University, Ankara, Turkey
Email address : [email protected] Burcu Ungor, Department of Mathematics, Ankara University, Ankara, Turkey
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